If one adds an inductive subset to a model of $ACA_0$, do we always get a new model of $ACA_0$? Suppose $(M, \mathcal X) \models ACA_0$. Recall that a subset $A \subseteq M$ is $inductive$ over $M$ if $M$ satisfies all instances of induction in the expanded language with a predicate for $A$. Suppose $A$ is inductive and let us denote by $(M, \mathcal X[A])$ the model whose second order part consists of all sets definable from parameters in $\mathcal X$ and $A$. What I want to know is whether or not this is necessarily again a model of $ACA_0$.
If it helps, my intended application concerns countable models ($M$ and $\mathcal X$ are both countable) and I know that the $A$ I have in mind is in fact inductive in the larger language i.e. $M$ satisfies all instances of induction in the language with predicates for $A$ and all elements of $\mathcal X$ (note that every set in a model of $ACA_0$ is inductive). 
I suppose more generally I'm interested in knowing, given $A \notin \mathcal X$ which is inductive over $M$, what do I need to require of a $\mathcal Y$ so that $A \in \mathcal Y$, $\mathcal X \subseteq \mathcal Y$ and $(M, \mathcal Y) \models ACA_0$?
 A: The answer to your (first) question is in the negative.
More explicitly: given any countable nonstandard model $M$ of PA, there are inductive subsets $A$ and $B$ of $M$ such that the expansion $(M,A,B)$ fails to satisfy induction in the extended language. I know two ways to achieve this:
(1) Use (Cohen) forcing to construct $A$ and $B$. This was first done by Andrzej Mostowski in his paper: A remark on models of the Gödel-Bernays axioms for set theory. Sets and classes (on the work by Paul Bernays), pp. 325–340. Studies in Logic and the Foundations of Math., Vol. 84, North-Holland, Amsterdam, 1976. 
Mostowski's strategy is similar to the proof of the theorem credited to Hugh Woodin in this MO answer of Joel Hamkins.
(2) Use tools of model theory (Beth's theorem, or that of Svenonius) to show that there are countable nonstandard models of PA (or ZF) that carry distinct (indeed continuum many) full inductive satisfaction classes. It is easy to see that if $A$ and $B$ are full satisfaction classes over a model $M$ of PA such that the expansion $(M,A,B)$ satisfies PA in the extended language, then $A=B$, so if $A$ and $B$ are distinct full inductive satisfaction classes over a model $M$ of PA (or ZF), then the expansion $(M,A,B)$ does not satisfy PA (or ZF) in the extended language.
So $A$ and $B$ are as in (1) or (2) above and $\cal{X}$ is chosen as the collection of all subsets of $M$ that are definable in $(M,A)$, then the model $(M,\cal{X}$$[B]$) fails to satisfy $ACA_0$ even though $(M,\cal{X})$ does satisfy $ACA_0$ and $B$ is inductive.
As for the second question: the only necessary and sufficient condition for $(M,\cal{X}$$[A]$) to satisfy $ACA_0$  that I know of is that $(M,X,A)$ satisfies PA in the extended language for every $X \in \cal{X}$.
